Q17 of 22 Page 1

Conductivity of 2.5 10-4M methanoic acid is 5.25 10-5 Scm-1. Calculate its molar conductivity and degree of dissociation. Given: (H+) = 349.5 Scm2mol-1 and ⋀°(HCOO-) =50.5 Scm2mol-1.

Given, K, conductivity of HCOOH = 5.25 10-5 Scm-1,


Molarity of HCOOH, M = 2.5 10-4M,


(H+) = 349.5 Scm2mol-1 and


⋀°(HCOO-) =50.5 Scm2mol-1


Molar conductance, ⋀m = Scm2mol-1


⋀m = = 210 Scm2mol-1


According to Kohlrauch’s Law, limiting molar conductivity of an electrolyte can be represented as sum of individual contribution of cation and anion of electrolyte.


Therefore, limiting molar conductivity of HCOOH can be written as,


⋀°m (HCOOH) = ⋀°m (HCOO-) + ⋀°m (H+)


= 50.5 Scm2mol-1 + 349.5 Scm2mol-1


= 400 Scm2mol-1


Degree of dissociation,


= 0.525


Molar conductivity and degree of dissociation of methanoic acid is 210 Scm2mol-1 and 0.525 respectively.


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(i) Phenol is more acidic than methanol.


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(iii) (CH3)3C – O – CH3 on reaction with HI gives (CH3)3C - I and CH3 – OH as the main products and not (CH3)3C – OH and CH3- I.